CHAPTER 3

ENERGY METHODS

3.0. INTRODUCTION

In the previous course, the relations existing between forces and deformations under various

loading conditions were considered. The analysis was based on two fundamental concepts,

the concept of stress and the concept of strain. And now third important concept to be

introduced is the concept of strain energy.

Strain energy of a member is defined as the increase in energy associated with the

deformation of the member. The view that the strain energy is equal to the work done by a

slowly increasing load applied to the member will be discussed. The strain-energy density of

a material will be defined as the strain energy per unit volume and we shall see that it is equal

to the area under the stress-strain curve of the material. From the stress-strain diagram of a

material we shall also define the modulus of toughness and modulus of resilience of the

material.

The elastic strain energy associated with normal stresses will be discussed, first in members

under axial loading and then in members in bending. Later we shall consider the elastic

strain energy associated with shearing stresses such as in torsional loadings of shafts and in

transverse loadings of beams. Strain energy for a general state of stress will be considered

where we shall derive the maximum-distortion-energy criterion for yielding.

3.1. STRAIN ENERGY

Consider a rod BC of length L and uniform cross-sectional area A, which is attached at B to a

fixed support, and subjected at C to a slowly increasing axial load P (Fig. 2.1). By plotting

the magnitude P of the load against the deformation x of the rod, we obtain a certain load-

deformation diagram (Fig. 2.2) which is characteristic of the rod BC.

Fig.3.1

Fig.3.2

Let us now consider the work dU done by the load P as the rod elongates by a small amount

dx. This elementary work is equal to the product of the magnitude P of the load and of the

smallelongation dx. We write

dU = P dx …(1)

and note that the expression obtained is equal to the element of area of

width dx located under the load-deformation diagram (Fig. 3.3). The

total work U done by the load as the rod undergoes a deformation x1 is

thus

and is equal to the area under the load-deformation diagram between x = 0 and x = x1.

The work done by the load P as it is slowly applied to the rod must result in the increase of

some energy associated with the deformation of the rod. This energy is referred to as the

strain energy of the rod. We have, by definition,

Strain energy = U = 1

0

x

dxP …(2)

We recall that work and energy should be expressed in units obtained by multiplying units of

length by units of force. Thus, if SI metric units are used, work and energy are expressed in

N.m; this unit is called a joule (J).

In the case of a linear and elastic deformation, the portion of the

load-deformation diagram involved may be represented by a straight lineof

equation P = kx (Fig. 2.4). Substituting for P in Eq. (2), we have

or

U = 112

1xP …(3)

where P1 is the value of the load corresponding to the deformation x1.

Fig.3.3

Fig. 3.4

The concept of strain energy is particularly useful in the

determination of the effects of impact loadings on

structures or machine components. Consider, for

example, a body of mass m moving with a velocity V0

which strikes the end B of a rod AB (Fig. 2.5a).

Neglecting the inertia of the elements of the rod, and

assuming no dissipation of energy during the impact, we

find that the maximum strain energy Um acquired by the rod (Fig. 2.5b) is equal to the

original kinetic energy of the moving body.

T = 2

02

1mV

We may then determine the value Pm of the static load which would have produced the same

strain energy in the rod, and obtain the value σm of the largest stress occurring in the rod by

dividing Pm by the cross-sectional area of the rod.

3.2. STRAIN-ENERGY DENSITY

We know that the load-deformation diagram for a rod BC depends upon the length L and the

cross-sectional area A of the rod. The strain energy U defined by Eq. (2), therefore, will also

depend upon the dimensions of the rod. In order to eliminate the effect of size from our

discussion and direct our attention to the properties of the material, we shall consider the

strain energy per unit volume. Dividing the strain energy U by the volume V = AL of the rod

(Fig. 2.1), and using Eq. (2), we have

Recalling that AP represents the normal stress σx in the rod, and Lx

the normal stain , we write

1

0

xxdV

U

Fig. 2.5

Fig. 3.5

where1 denotes the value of the strain corresponding to the elongation x1. The strain energy

per unit volume, VU , is referred to as the strain-energy density and will be denoted by the

letter u. We have, therefore,

Strain-energy density = u = xxd

1

0…(4)

The strain-energy density u is expressed in units obtained by dividing units of energy by units

of volume. Thus the strain-energy density is expressed in 3mJ or its multiples 3mkJ and

3mMJ .

Referring to (Fig. 2.6), we note that the strain-energy density u is equal to the area under the

stress-strain curve, measured from 0x to x = 1 . If the material is unloaded, the stress

returns to zero, but there is a permanent deformation represented by the strain p , and only

the portion of the strain energy per unit volume corresponding to the triangular area may be

recovered. The remainder of the energy spent in deforming the material is dissipated in the

form of heat. The value of the strain-energy density obtained by setting 1 =

R in Eq. (4),

where R is the strain at rupture, is known as the modulus of toughness of the material. It is

equal to the area under the entire stress-strain diagram (Fig. 2.7) and represents the energy

per unit volume required tocause the material to rupture. It is clear that the toughness of

a material is related to its ductility as well as to its ultimate strength, and that the capacity of a

structure to withstand an impact load depends upon the toughness of the material used.

If the stress σx remains within the proportional limit of the material, Hooke’s law applies and

we may write

σx = E x …(5)

Substituting for σx from (5) into (4), we have

Fig. 3.6 Fig. 3.7

u = 2

2

1

0

1

EdE xx …(6)

or, using Eq. (5) to express 1 in terms of the corresponding stress σ1 ,

Eu

2

2

1 …(7)

The value uy of the strain-energy density obtained by setting σ1 = σy in Eq. (7), where σy is

the yield strength, is called the modulus of resilience of the material. We have

yu = E

y

2

2…(8)

The modulus of resilience is equal to the area under the straight-

line portion OY of the stress-strain diagram (Fig. 2.8) and

represents the energy per unit volume that the material may

absorb without yielding. The capacity of a structure to

withstand an impact load without being permanently deformed

clearly depends upon the resilience of the material used.

Since the modulus of toughness and the modulus of resilience represent characteristic values

of the strain-energy density of the material considered, they are both expressed in J/m3 or its

multiples.

3.3. ELASTIC STRAIN ENERGY FOR NORMAL STRESSES

Since the rod considered in the preceding section was subjected to uniformly distributed

stresses σx, the strain-energy density was constant throughout the rod and could be defined as

the ratio VU of the strain energy U and the volume V of the rod. In a structural element or

machine part with a non-uniform stress distribution, the strain-energy density u may be

defined by considering the strain energy of a small element of material of volume V and

writing

or

dV

dUu …(9)

The expression obtained for it in Sec. 2.2 in terms of σx and εx remains valid, i.e., we still

have

Fig. 3.8

x

xxdu

0

…(10)

but the stress σx, the strain εx, and the strain-energy density u will generally vary from point

to point.

For values of σx within the proportional limit, we may set σx = Eεx in Eq. (10) and write

EEu x

xxx

2

2

2

1

2

1

2

1 …(11)

The value of the strain energy U of a body subjected to uniaxial normal stresses may be

obtained by substituting for it from Eq. (11) into Eq. (9) and integrating both members. We

have

dVE

U x

2

2…(12)

The expression obtained is valid only for elastic deformations and is

referred to as the elastic strain energy of the body.

Strain Energy under Axial Loading: We know that when a rod is

subjected to a centric axial loading, the normal stresses σx may be

assumed uniformly distributed in any given transverse section. Denoting by A the area of the

section located at a distance x from the end B of the rod (Fig. 2.9), and by P the internal force

in that section, we write σx = P/A. Substituting for σx into Eq. (12), we have

or setting dV = A dx,

In the case of a rod of uniform cross section subjected at its

ends to equal and opposite forces of magnitude P (Fig. 2.10),

Eq. (13) yields

EXAMPLES

1) A rod consists of two portions BC and CD of the same material

and same length, but of different cross sections (Fig. 2.11).

…(13)

…(14)

Fig. 3.9

Fig. 3.10

Fig. 3.11

Determine the strain energy of the rod when it is subjected to a centric axial load P,

expressing the result in terms of P, L, E, the cross-sectional area A of portion CD, and the

ratio n of the two diameters.

Solution

We use Eq. (14) to compute the strain energy of each of the two portions, and add the

expressions obtained:

or

We check that, for n = 1, we have

which is the expression given in Eq. (14) for a rod of length L and uniform cross section of

area A. We also note that, for n > 1, we have Un< U for example, when n = 2, we have U2=

18

5U

. Since the maximum stress occurs in portion CD of the rod and is equal to σmax. =

P/A, it follows that, for a given allowable stress, increasing the diameter of portion BC of the

rod results in a decrease of the overall energy-absorbing capacity of the rod. Unnecessary

changes in cross-sectional area should therefore be avoided in the design of members which

may be subjected to loadings, such as impact loadings, where the energy-absorbing capacity

of the member is critical.

2) A load P is supported at B by two rods of the same material and of the same uniform cross

section of area A (Fig. 3.12). Determine the strain energy of the system.

Solution

…(15)

Denoting by FBC and FBD, respectively, the forces in members BC and BD, and recalling Eq.

(14), we express the strain energy of the system as

AE

BDF

AE

BCFU BDBC

22

22

…(16)

But we note from Fig. 3.12 that

lBDlBC 8.06.0

and from the free-body diagram of pin B and the corresponding force triangle (Fig. 3.13) that

Substituting into Eq. (16), we have

Strain Energy in Bending: Consider a beam AB subjected to a given loading (Fig. 3.14), and

let M be the bending moment at a distance x from end A . Neglecting for the time being the

effect of shear, and taking into account only the normal stresses σx

= IMy , we substitute this expression into Eq. (12) and write

Fig. 3.12 Fig. 3.13 D

Fig. 3.14

Setting dv = dA dx, where dA represents an element of the cross- sectional area, and recalling

that M2/2EI

2 is a function of x alone, we have

Recalling that the integral within the parentheses represents the moment of inertia I of the

cross section about its neutral axis, we write

3) Determine the strain energy of the prismatic cantilever beam AB (Fig. 3.15), taking into

account only the effect of the normal stresses.

Solution

The bending moment at a distance x from end A is M = -Px .

Substituting this expression into Eq. (17), we write

3.4. ELASTIC STRAIN ENERGY FOR SHEARING STRESSES

When a material is subjected to plane shearing stresses the strain-energy

density at a given point may be expressed as

where xy is the shearing strain corresponding to xy (Fig. 3.16a). We

note that the strain-energy density u is equal to the area under the

shearing stress-strain diagram (Fig. 2. 16b).

…(17)

Fig.3.15

…(18)

Fig. 3.16

For values of xy within the proportional limit, we have xy = G xy , where G is the modulus of

rigidity of the material. Substituting for xy into Eq. (18) and performing the integration, we

write

The value of the strain energy U of a body subjected to plane shearing stresses may be

obtained by recalling from Sec. 3.4 that

Substituting for it from Eq. (19) into Eq. (9) and integrating both members, we have

This expression defines the elastic strain associated with the shear

deformations of the body. As the similar expression obtained in Sec. 3.4 for

uniaxial normal stresses, it is valid only for elastic deformations.

Strain Energy in Torsion: Consider a shaft BC of length L subjected to

one or several twisting couples. Denoting by J the polar moment of inertia

of the cross section located at a distance x from B (Fig. 3.17), and by T the internal torque in

that section, we recall that the shearing stresses in the section are xy = JT . Substituting for

xy into Eq. (20), we have

Setting dv = dA dx, where dA represents an element of the cross- sectional area, and

observing that T2/2GJ

2 is a function of x alone, we write

…(19)

…(9)

…(20)

Fig. 3.17

Recalling that the integral within the parentheses represents the polar moment of inertia J of

the cross section, we have

In the case of a shaft of uniform cross section subjected at its ends

to equal and opposite couples of magnitude T (Fig. 3.18), Eq. (21)

yields

4) A circular shaft consists of two portions BC and CD of the same

material and same length, but of different cross sections (Fig. 3.19).

Determine the strain energy of the shaft when it is subjected to a

twisting couple T at end D, expressing the result in terms of T, L, G,

the polar moment of inertia J of the smaller cross section, and the

ratio a of the two diameters.

Solution

We use Eq. (22) to compute the strain energy of each of the two portions of shaft, and add the

expressions obtained. Noting that the polar moment of inertia of portion BC is equal to n4J,

we write

or

We check that, for n = 1, we have

… (21)

… (22) Fig. 3.18

Fig. 3.19

…(23)

which is the expression given in Eq. (22) for a shaft of length L and uniform cross section.

We also note that, for n > 1, we have Un<U1 ; for example, when n = 2, we have U2 =

132

17U

.

Since the maximum shearing stress occurs in the portion CD of the shaft and is proportional

to the torque T, we note as we did earlier in the case of the axial loading of a rod that, for a

givenallowable stress, increasing the diameter of portion BC of the shaft results in a decrease

of the overall energy-absorbing capacity of the shaft.

Strain Energy under Transverse Loading: In Sec. 3.4 we obtained an expression for the

strain energy of a beam subjected to a transverse loading. However, in deriving that

expression we took into account only the effect of the normal stresses due to

bending and neglected theeffect of the shearing stresses. We shall now take

into account the effect of both types of stresses.

5) Determine the strain energy of the rectangular cantilever beam AB (Fig.

3.20), taking into account the effect of both normal and shear stresses.

Solution

We first recall from Example 3 that the strain energy due to the normal stresses σx is

To determine the strain energy Uτ due to the shearing stresses we apply the equation below

for a beam with a rectangular cross section of width b and depth h,

Substituting for xy into Eq. (20), we write

Fig. 3.20

or, setting dV = b dy dx, and after reductions,

Performing the integrations, and recalling that c = h/2, we have

The total strain energy of the beam is thus or, noting that I/A = h

2/12 and factoring the expression for Uσ,

We know that 3EG , we conclude that the parenthesis in the expression obtained is less

than 29.01 Lh and, thus, that the relative error is less than 29.0 Lh when the effect of

shear is neglected. For a beam with a ratio Lh less than 101 , the percentage error is less

than 0.9%. It is therefore customary in engineering practice to neglect the effect of shear in

computing the strain energy of slender beams.

3.5. STRAIN ENERGY FOR A GENERAL STATE OF STRESS

In the preceding sections, we determined the strain energy of a body in a state of uniaxial

stress (Sec. 2.3) and in a state of plane shearing stress (Sec. 2.4). In the case of a body in a

general state of stress characterized by the six stress components ,x ,y ,z zxyzxy and ,,

, the strain-energy density may be obtained by adding the expressions given in Eqs. (10) and

(18), as well as the four other expressions obtained through a permutation of the subscripts.

The strain energy density of a three-dimensional principal stress system is given by:

…(24)

…(25)

where σa, σb, and σc are the principal stresses at the given point.

This total strain energy can be conveniently considered as made up of two parts:

(a) the volumetric or dilatational strain energy density;

(b) the shear or distortional strain energy density.

Therefore we write

(a) The volumetric or dilatational strain energy density

This is the strain energy density associated with a mean or hydrostatic stress of

cba3

1acting equally in all three mutually perpendicular directions giving rise

to no distortion, merely a change in volume. And we can set also

Volumetric strain energy density (uv ) =

2

6

21cba

E

…(26)

(b) The shear or distortional strain energy density

In order to consider the general principal stress case it is necessary, to add to the mean stress

in the three perpendicular directions, certain so-called deviatoric stress values to return the

stress system to values of a , b and c . These deviatoric stresses are then associated directly

with change of shape, i.e. distortion, without change in volume and the strain energy

associated with this mechanism is given by

In the case of plane stress, and assuming that the c axis is perpendicular to

Fig. 3.21

the plane of stress, we have σc = 0 and Eq. (27) reduces to

EXAMPLES

6) During a routine manufacturing operation, rod AB must acquire

an elastic strain energy of 12J. Using E = 200 GPa, determine the

required yield strength of the steel if the factor of safety with

respect to permanent deformation is to be five.

Design strain energy : Since a factor of safety of five is required, the rod should be designed

for a strain energy of

Strain-Energy Density: The volume of the rod is

Since the rod is of uniform cross section, the required strain-energy density is

Yield Strength: We recall that the modulus of resilience is equal

to the strain-energy density when the maximum stress is equal to

σy. Using Eq. (8),we write

σy = 250.8 MPa = 251

…(27)

Comment

It is important to note that, since energy loads are not linearly

related to the stresses they produce, factors of safety associated with

energy loads should be applied to the energy loads and not to the

stresses.

7) (a) Taking into account only the effect of normal stresses due to bending, determine the

strain energy of the prismatic beam AB for the loading shown. (b) Evaluate the strain energy,

knowing that the beam is a W 250 x 67, P = 180 kN, L = 3.6 m, a = 0.9 m, b = 2.7 m and E =

200 GPa.

Bending Moment : Using the free-body diagram of the entire beam, we determine the

reactions

For portion AD of the beam, the bending moment is

For portion DB, we note that the bending moment at a distance v from end B is

(a) Strain Energy: Since strain energy is a scalar quantity, we add

the strain energy of portion AD to that of portion DB to obtain the

total strain energy of the beam. Using Eq. (17), we write

or, since (a + b) = L,

(b) Evaluation of the Strain Energy: The moment of inertia of a W 250 x 67 rolled-steel

shape is obtained from Appendix C and the given data is repeated here:

Substituting into the expression for U, we have

U = 429 J

3.6. IMPACT LOADING

Consider a rod BD of uniform cross section which is hit at its end B by a body of mass m

moving with a velocity vo (Fig. 3.22a). As the rod deforms under

the impact (Fig. 3.22b), stresses develop within the rod and reach a

maximum value σm. After vibrating for a while, the rod will come

to rest, and all stresses will disappear. Such a sequence of events is

referred to as an impact loading.

In order to determine the maximum value σm of the stress

occurring at a given point of a structure subjected to an impact

loading, we shall make several simplifying assumptions.

First, we shall assume that the kinetic energy T = 2

2

1omV of the striking body is transferred

entirely to the structure and, thus, that the strain energy Um corresponding to the maximum

deformation xm is

…(28)

Fig. 3.22

This assumption leads to the following two specific requirements:

1. No energy should be dissipated during the impact.

2. The striking body should not bounce off the structure and retain part of its energy. This, in

turn, necessitates that the inertia of the structure be negligible, compared to the inertia of the

striking body.

In practice, neither of these requirements is satisfied, and only part of the kinetic energy of

the striking body is actually transferred to the structure. Thus, assuming that all of the kinetic

energy of the striking body is transferred to the structure leads to a conservative design of that

structure.

We shall further assume that the stress-strain diagram obtained from a static test of the

material is also valid under impact loading. Thus, for an elastic deformation of the structure,

we may express the maximum value of the strain energy as

In the case of the uniform rod of Fig. 3.22, the maximum stress σm has the same

value throughout the rod, and we may write Um= EVm 22 . Solving for σm and substituting

for Um from Eq.(28), we write

We note from the expression obtained that selecting a rod with a large volume V and a low

modulus of elasticity E will result in a smaller value of the maximum stress σm for a given

impact loading.

In most problems, the distribution of stresses in the structure is not uniform, and formula (30)

does not apply. It is then convenient to determine the static load Pm which would produce the

same strain energy as the impact loading, and compute from Pm the corresponding value σm

of the largest stress occurring in the structure.

EXAMPLES

8) A body of mass m moving with a velocity vo hits the end B of

the non-uniform rod BCD (Fig. 3.23). Knowing that the diameter

of portion BC is twice the diameter of portion CD, determine the

maximum value σm of the stress in the rod.

…(29)

…(30)

Fig.3.23

Solution

Making n = 2 in the expression (15) obtained in Example 1, we find that when rod BCD is

subjected to a static load Pm its strain energy is

where A is the cross-sectional area of portion CD of the rod.

Solving Eq. (31) for Pm we find that the static load which produces in the rod the same strain

energy as the given impact loading is

where Um is given by Eq. (28). The largest stress occurs in portion CD of the rod. Dividing

Pm by the area A of that portion, we have

or, substituting for Um from Eq. (28),

Comparing this value with the value obtained for Um in the case of the uniform rod of Fig.

3.22 and making V = AL in Eq. (30), we note that the maximum stress in the rod of variable

cross section is 26.5% larger than in the lighter uniform rod. Thus, as we observed earlier in

our discussion of Example 1, increasing the diameter of portion BC of the rod results in a

decrease of the energy-absorbing capacity of the rod.

9) A block of weight W is dropped from a height h onto the free end of

the cantilever beam AB (Fig. 3.24). Determine the maximum value of

the stress in the beam.

…(31)

…(32)

Fig.3.24

Solution

As it falls through the distance h, the potential energy Wh of the block is transformed into

kinetic energy. As a result of the impact, the kinetic energy in turn is transformed into strain

energy. We have, therefore,

Recalling the expression obtained for the strain energy of the cantilever beam AB in Example

3 and neglecting the effect of shear, we write

Solving this equation for Pm, we find that the static force which produces in the beam the

same strain energy is

The maximum stress σm occurs at the fixed end B and is equal to

Substituting for Pm from (34), we write

or, recalling (33),

3.7. DESIGN FOR IMPACT LOADS

We shall now compare the values obtained in the preceding section for the maximum stress

σm (a) in the rod of uniform cross section of Fig. 3.22, (b) in the rod of variable cross section

of Example 8, and (c) in the cantilever beam of Example 9, assuming that the latter has a

circular cross section of radius c.

…(33)

…(34)

…(35)

(a) We first recall from Eq. (30) that, if Um denotes the amount of energy transferred to the

rod as a result of the impact loading, the maximum stress in the rod of uniform cross section

is

where V is the volume of the rod.

(b) Considering next the rod of Example 8 and observing that the volume of the rod is

we substitute AL = 2V/5 into Eq. (32) and write

c) Finally, recalling that I = 4

4

1c for a beam of circular cross section, we note that

where V denotes the volume of the beam. Substituting into Eq. (35), we express the

maximum stress in the cantilever beam of Example 9 as

We note that, in each case, the maximum stress σm is proportional to the square root of the

modulus of elasticity of the material and inversely proportional to the square root of the

volume of the member. Assuming all three members to have the same volume and to be of

the same material, we also note that, for a given value of the absorbed energy, the uniform

rod will experience the lowest maximum stress, and the cantilever beam the highest one.

This observation may be explained by the fact that, the distribution of stresses being uniform

in case (a), the strain energy will be uniformly distributed throughout the rod. In case (b), on

the other hand, the stresses in portion BC of the rod are only 25% as large as the stresses in

portion CD. This uneven distribution of the stresses and of the strain energy results in a

…(36a)

…(36b)

…(36c)

maximum stress σm twice as large as the corresponding stress in the uniform rod. Finally, in

case (c), where the cantilever beam is subjected to a transverse impact loading, the stresses

vary linearly along the beam as well as across a transverse section. The very uneven resulting

distribution of strain energy causes the maximum stress σm to be 3.46 times larger than if the

same member had been loaded axially as in case (a).

The properties noted in the three specific cases discussed in this section are quite general and

may be observed in all types of structures and impact loadings. We thus conclude that a

structure designed to withstand effectively an impact load should

1. Have a large volume

2. Be made of a material with a low modulus of elasticity and a high yield strength

3. Be shaped so that the stresses are distributed as evenly as possible throughout the structure 3.8. WORK AND ENERGY UNDER A SINGLE LOAD

When we first introduced the concept of strain energy at the beginning of this chapter, we

considered the work done by an axial load P applied to the end of a rod of uniform cross

section (Fig. 2.1). We defined the strain energy of the rod for an elongation x1 as the work of

the load P as it is slowly increased from 0 to the value P1 corresponding to x1. We wrote

In the case of an elastic deformation, the work of the load P, and thus the strain energy of the

rod, were expressed as

However, when a structure or member is subjected to a single

concentrated load, it is possible to use Eq. (3) to evaluate its elastic

strain energy, provided that the relation between the load and the

resulting deformation is known. For instance, in the case of the

cantilever beam of Example 3 (Fig. 3.25), we write

…(2)

…(3)

Fig. 3.25

and, substituting for y the value obtained from the table of Beam Deflections and Slopes

we have

A similar approach may be used to determine the strain energy of a

structure or member subjected to a single couple. Recalling that the

elementary work of a couple of moment M is M dθ, where dθ is a

small angle. Since M and θ are linearly related the elastic strain energy of a cantilever beam

AB subjected to a single couple M1 at its end A (Fig. 3.26) may be expressed as

where θ1 is the slope of the beam at A. Substituting for θ1 the value

obtained fromthe table of Beam Deflections and Slopes, we write

In a similar way, the elastic strain energy of a uniform circular

shaftAB of length L subjected at its end B to a single torque T1 (Fig.

3.27)may be expressed as

Substituting for the angle of twist JG

LT11 , we verify that

…(37)

…(38)

Fig.3.26

…(39)

…(40)

Fig.3.27

Example

10.A block of mass m moving with a velocity Vo hits squarely the prismatic member AB at

its midpoint C (Fig. 3.28). Determine (a) the equivalent static load Pm, (b) the maximum

stress σm in the member, and (c) the maximum deflection xm at point C.

(a) Equivalent Static Load: The maximum strain energy of the member is equal to the

kinetic energy of the block before impact. We have

On the other hand, expressing Um as the work of the equivalent horizontal static load as it

is slowly applied at the midpoint C of the member, we write

where xm is the deflection of C corresponding to the static load Pm. From the table of Beam

Deflections and Slopes , we find that

Substituting for xm from (43) into (42), we write

Solving for Pm and recalling Eq. (41), we find that the static load equivalent to the given

impact loading is

…(41)

…(42)

…(43)

…(44)

Fig. 3.28 Fig. 3.29

(b) Maximum Stress: Drawing the free-body diagram of the member (Fig. 3.29), we find

that the maximum value of the bending moment occurs at C and is Mmax = PmL/4. The

maximum stress, therefore, occurs in a transverse section through C and is equal to

Substituting for Pm from (44), we write

(c) Maximum Deflection: Substituting into Eq. (43) the expression obtained for Pm in (44),

we have

3.9. DEFLECTION UNDER A SINGLE LOAD BY THE WORK-ENERGY METHOD

We saw in the preceding section that, if the deflection x1 of a structure or member under a

single concentrated load P1 is known, the corresponding strain energy U may be obtained by

writing

A similar expression may be used to obtain the strain energy of a structural member under a

single couple M1

Conversely, if the strain energy U of a structure or member subjected to a single concentrated

load P1 or couple M1 is known, Eq. (3) or (38) may be used to determine the corresponding

deflection x1 or angle θ1. In order to determine the deflection under a single load applied to a

structure consisting of several component parts, we may find it easier, to first compute the

strain energy of the structure by integrating the strain-energy density over its various parts, as

was done in Secs. 3.4 and 3.5, and then use either

Eq. (3) or Eq. (38) to obtain the desired deflection. Similarly, the angle of twist of a

composite shaft may be obtained by integrating the strain-energy density over the various

parts of the shaft and solving Eq. (40) for 1 .

…(3)

…(38)

It should be kept in mind that the method presented in this section may be used only if the

given structure is subjected to a single concentrated load or couple.

Example 11. A load P is supported at B by two uniform rods of the same cross-sectional area A (Fig.

3.30). Determine the verticaldeflection of point B.

Solution

The strain energy of the systemunder the given load was determined in

Example 2. Equating the expression obtained for U to the work of the

load, we write

and, solving for the vertical deflection of B,

12.Determine the deflection of end A of the cantilever beam AB

(Fig. 3.31), taking into account the effect of(a) the normal

stresses only, (b) both the normal and shearing stresses.

Solution

(a) Effect of Normal Stresses: The work of the force P as it is slowly applied to A is

Substituting for U the expression obtained for the strain energy of the beam in Example 3,

where only the effect of thenormal stresses was considered, we write

and, solving for yA,

Fig. 3.30

Fig. 3.31

(b) Effect of Normal and Shearing Stresses : We now substitute for U the expression (24)

obtained in Example 5, where the effects of both the normal and shearingstresses were taken

into account. We have

(…24) ;

and, solving for yA,

We note that the relative error when the effect of shear is neglected is the same that was

obtained in Example 5, i.e., less than 0.9(h/L)2. As we indicated then, this is less than 0.9%

for a beam with a ratio h/L less than 1/10.

12.A torque T is applied at the end D of shaft BCD (Fig. 2.32). Knowing that both portions of

the shaft are of the same material and same length, but that the diameter of BC is twice the

diameter of CD, determine the angle of twist for the entire shaft.

Solution

The strain energy of a similar shaft was determined in

Example 4 by breaking the shaft into its component

parts BC and CD. Making n = 2 in Eq. (23), we have

(…23) ;

where G is the modulus of rigidity of the material, and J the polar moment of inertia of

portion CD of the shaft. Setting U equal to the work of the torque as it is slowly applied and

recalling Eq. (40), to end D, we write

(…40) ;

and, solving for the angle of twist BD ,

v

Fig. 2.32 Fig. 3.32

Sample Problems

1.The block D of mass m is released from rest and falls a

distance h before it strikes the midpoint C of the aluminum

beam AB. Using E = 70 GPa, determine (a) the maximum

deflection of point C, (b) the maximum stress which occurs in the beam.

Solution Principle of Work and Energy: Since the block is released

from rest, we note that in position1 both the kinetic energy and

the strain energy are zero. Inposition 2, where the maximum

deflection ym occurs, the kinetic energy is again zero. Referring

to the table of Beam Defections and Slopes we find the

expression for ym shown. The strain energy of the beam in

position 2 is

We observe that the work done by the weight W of the block is W(h + ym).

Equating the strain energy of the beam to the work done by W, we have

(a) Maximum Deflection of Point C: From the given data we have

Substituting into Eq. (1), we have

Solving this quadratic equation, we find

…(1)

(b) Maximum Stress: The value of Pm is

Since Mmax = LPm4

1, the maximum stress is

2.Members of the truss shown consist of sections of

aluminum pipe with the cross- sectional areas indicated.

Using E = 70 GPa, determine the vertical deflection

of point E caused by the load P.

Axial Forces in Truss Members: The reactions are found by

using the free- body diagram of the entire truss. We then consider in sequence the equilibrium

of joints, E, C, D, and B. At each joint we determine the forces indicated by dashed lines. At

joint B, the equation 0 xF provides a check of our computations.

Strain Energy: Noting that E is the same for all members, we express the strain energy of

the truss as follows

where Fi is the force in a given member as indicated in the following table and where the

summation is extended over all members of the truss.

Returning to Eq. (1), we have U = (1/2E)(29.7 X 103P2). Principle of Work-Energy: We recall that the work done by the load P as it is gradually

applied is EPy2

1. Equating the work done by P to the strain energy U and recalling that E =

70 GPa and P = 40 kN, we have

or

3.10. WORK AND ENERGY UNDER SEVERAL LOADS

In this section, we shall see how the strain energy of a structure subjected to several loads

may be expressed in terms of the loads and the resulting

deflections.

Consider an elastic beam AB subjected to two concentrated

loads F1 and F2. The strain energy of the beam is equal to the

work of P1 and P2as they are slowly applied to the beam at C1

and C2, respectively (Fig. 3.33). However, in order to evaluate this work, we must first

express the deflections x1 and x2 in terms of the loads P1 and P2.

Let us assume that only P1 is applied to the beam (Fig.

3.34). We note that both C1 and C2 are deflected and that

Fig. 3.33

Fig. 3.34

their deflections areproportional to the load P1. Denoting these deflections by x11 and

x21,respectively, we write

where α11 and α21 are constants called influence

coefficients. These constants represent the

deflections of C1 and C2, respectively, when a unit

load is applied at C1 and are characteristics of the

beam AB.

Let us now assume that only P2 is applied to the beam (Fig. 3.35).

Denoting by x12 and x22, respectively, the resulting deflections of C1 andC2, we write

where α12 and α22 are the influence coefficients representing the deflections of C1 and C2,

respectively, when a unit load is applied at C2. Applying the principle of superposition, we

express the deflections x1 and x2 of C1 and C2 when both loads are applied (Fig. 3.33) as

To compute the work done by P1 and P2, and thus the strain energy of the beam, we shall

find it convenient to assume that P1 is first applied slowly at C1 (Fig. 3.36a).

Recalling the first of Eqs. (3.41), we expressthe work of P1 as

…(41)

…(42)

…(43)

…(44)

Fig. 3.35

…(45)

and note that P2 does no work while C2 moves through x21, since it has not yet been applied to

the beam.

Now we slowly apply P2 at C2 (Fig. 2.36b); recalling the second of Eqs. (42), we express the

work of P2 as

But, as P2 is slowly applied at C2, the point of application of P1

moves through from x12, from '

1C to C1,and the load P1 does

work. Since P1 is fully applied during this displacement (Fig,

3.37), its work is equal to P1x12 or, recalling the first of Eqs.

(42),

Adding the expressions obtained in (45), (46), and (47), we express the strain energy of the

beam under the loads P1 and P2 as

If the load P2 had first been applied to the beam (Fig. 3.38a), and then the load P1 (Fig.

3.38b), the work done

by each load would

have been as shown

in Fig. 3.39.

…(46)

…(47)

…(48)

Fig 3.36

2.37 Fig. 3.37

Calculations similar to those we have just carried out would lead to the following alternative

expression for the strain energy of the beam:

Equating the right-hand members of Eqs. (48) and (49), we

find that α12= α21, and thus conclude that the deflection

produced at C1 by a unit load applied at C2 is equal to the

deflection produced at C2 by a unit load applied at C1. This

is known as Maxwell’s reciprocal theorem, after the British

physicist James Clerk Maxwell (1831-1879).

While we are now able to express the strain energy U of a

structure subjected to several loads as a function of these

loads, we cannot use the method of Sec. 3.9 to determine

the deflection of such a structure. Indeed, computing the

strain energy U by integrating the strain-energy density u over the structure and substituting

the expression obtained into (48) would yield only one equation, which clearly could not be

solved for the various coefficients α.

3.11. CASTIGLIANO’S THEOREM

We recall the expression obtained in the preceding section for the strain energy of an elastic

structure subjected to two loads P1 and P2:

…(49)

Fig. 3.39

Fig. 3.38

…(50)

Fig. 3.38

where α11, α12 and α22 are the influence coefficients associated with the points of application

C1 and C2 of the two loads. Differentiating both members of Eq. (48) with respect to P1 and

recalling Eq. (43), we write

Differentiating both members of Eq. (50) with respect to P2, recalling Eq. (44), and keeping

in mind that α12 = α21, we have

More generally, if an elastic structure is subjected to n loads P1, P2,…, Pn the deflection of the

point of application of fP , measured alongthe line of action of fP , may be expressed as the

partial derivative of the strain energy of the structure with respect to the load P. We write

This is Castigliano’s theorem, named after the Italian engineer AlbertoCastigliano (1847-

1884) who first stated it.

Recalling that the work of a couple M is M2

1, where is the angle of rotation at the point

where the couple is slowly applied, we note that Castigliano’s theorem may be used to

determine the slope of a beam at the point of application of a couple jM . We have

…(51)

…(52)

…(53)

…(54)

Similarly, the angle of twist j in a section of a shaft where a torque Tj is slowly applied is

obtained by differentiating the strain energy of the shaft with respect to :jT

3.11. DEFLECTIONS BY CASTIGLIANO’S THEOREM

We saw in the preceding section that the deflection jx of a structure at the point of application

of a load P may be determined by computing the partial derivative jPU of the strain

energy U of the structure. As we recall from Secs. 3.4 and 3.5, the strain energy U is obtained

by integrating or summing over the structure the strain energy of each element of the

structure. We shall find that the calculation by Castigliano’s theorem of the deflection jx is

simplified if the differentiation with respect to the load jP is carried out before the integration

or summation.

In the case of a beam, for example, we recall from Sec. 3.4 that

and determine the deflection jx of the point of application of the load jP by writing

In the case of a truss consisting of n uniform members of length iL , cross-sectional area Ai,

and internal force Fi, we recall Eq. (14) and express the strain energy U of the truss as

…(55)

…(56)

…(57)

…(58)

The deflection jx of the point of application of the load jP is obtained by differentiating with

respect to each term of sum. We write

EXERCISES

Strain Energy & Strain energy density

1. Determine the modulus of resilience for each of the following

metals:

(a) Stainless steel (cold-rolled): E = 190 GPa, sy = 520 MPa

(b) Stainless steel (annealed): E = 190 GPa, sy = 260 MPa

(c) Malleable cast iron: E = 165 GPa, sy = 230 MPa

2. The stress-strain diagram shown in Fig. 1 has been drawn from

data obtained during the tensile test of an aluminum alloy. Using E = 72 GPa, (a) determine

the modulus of resilience of the alloy, (b) determine by approximate means the modulus of

toughness of the alloy.

3. Using E = 75 GPa, determine (a) the strain energy of the aluminum rod ABC when

a = 1.2 m, (b) the corresponding strain-energy density in portions AB and BC of the rod.

(Refer Fig. 2 )

…(59)

Fig. 1

4. In the truss shown in Fig. 3 , all members are

made of the same material and have the uniform

cross-sectional areas indicated. Determine the

strain energy of the truss when the load P is

applied.

5. Assuming that the prismatic beam AB in Fig.4 has a

rectangular cross section, show that for the given loading the

maximum value of the strain-energy density in the beam is

where U is the strain energy of the beam and V is its volume.

Fig. 2

Fig. 3

Fig. 4

6. Two solid shafts are connected by

the gears shown in Fig. 5. Using G =

77 GPa , determine the strain energy

of each shaft when a 1.1 kN- m torque

T is applied at D.

7. Determine the strain energy of the prismatic beam AB shown in Fig. 6, taking into account

the effect of both normal and shearing stresses.

8. An aluminum tube having the cross section shown in Fig.7 is struck squarely in its

midsection by a 6-kg block moving horizontally with a speed of 2 m/s. Using E = 70 GPa,

determine (a) the equivalent static load, (b) the maximum stress in the beam, (c) the

maximum deflection at the midpoint C of the beam.

Fig.5

Fig. 6

Fig. 7

9. Using the method of work and energy, determine

the deflection at point D caused by the load P.

(Refer Fig. 8)

10. The 20-mm-diameter steel rod BC is attached to the lever AB and to the fixed support C

shown in Fig.9. The uniform steel lever is 10 mm thick and 30 mm deep. Using the method

of work and energy, determine the length L of rod BC for which the deflection of point A is

40 mm. Use E = 200 GPa and G = 77 GPa.

work and energy method

11. Using the method of work and energy, determine the

deflection at point D caused by the load P.

Ans.[ ]32 EILapa

12. Using the method of work and energy, determine the slope

at point D caused by the couple M0.

[Ans. EIaLM o 33 ]

13. In the assembly shown, member BC is a solid steel rod

of 25-mm diameter and all other members are made of 12.5

Fig. 8

Fig. 9

mm diameter steel rods. Using F 200 GPa, determine the deflection of point D caused by the

25-kN load.

[Ans. 1.333 mm ]

Castigliano’s theorem

14. For the uniform rod and loading shown and using Castigliano’s

theorem, determine (a) the horizontal deflection of point B, (b) the

vertical deflection of point B.

[Ans. a) EIPRbEIPR 4)2 33 ]

15. Three rods, each of the same flexural rigidity El, are welded to form

the frame ABCD. For the loading shown, determine (a) the deflection

of point D, (b) the angle formed by the frame at point D and the

vertical.

[Ans. (a) EIPLbEIPL 6)(;65 23 ]

16. Two rods of the same flexural rigidity El are welded at B.

For the loading shown, determine (a) the horizontal deflection of

point B, (b) the vertical deflection of point B.

[ Ans. Find yourself ]

REFERENCES

1. Beer, Ferdinand P. and Johnston E.R.I. , Mechanics of Materials, 4th

ed., 2005.

2. Gere, James M., Mechanics of materials, 6th

ed., 2003.